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July 20th, 2010, 10:23 PM   #1
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The Lagrange multiplier method

1. Use the Lagrange multiplier method to solve the following. [assume that the method works, i.e. that the critical point you find is the right kind of extremum]

(a) min: (x^2) + xy + (y^2)
subject to: x+y=10

(b) max: ?x + 2?y
subject to: 2x + y = 18

The solutions are:
(a) min=75 at (5,5)
(b) max=9 at (1,16)

But how to get to the solutions? Can someone solve and explain for me please?

2. Use the Lagrange multiplier method to find the closest and furthest points on the unit sphere in R^3 to the point (2; 2; 2). That is, use the Lagrange multiplier method on each of the two problems:
max/min : (||(2,2,2)-(x,y,z)||)^2
subject to: (||(x,y,z)||)^2 =1
Note: "||" above are absolute value brackets.
If you apply the method correctly, you should get two answers, one of which will be the minimum and one of which will be the maximum.

Can someone please solve this for me? At least I have an idea about 1. but this is ticking me off.

Thank you!
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July 21st, 2010, 05:25 AM   #2
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Re: The Lagrange multiplier method

It sounds like you need an introduction to the basics of Lagrange multipliers. See ... range.html
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